Official Answer: - pi

Cross Product:

\(\displaystyle

\int_{0}^{2\pi} \int_{0}^{1} ( r cos \theta , r sin \theta , -2 ) * ( 2r^2 cos \theta , 2r^2 sin \theta , r ) ~ dr d \theta

\)

\(\displaystyle

\int_{0}^{2\pi} \int_{0}^{1} ( 2r^3 - 2r ) ~dr d \theta = -\pi

\)

I managed to get the correct answer. but I didn't understand why the Jacobian of 'r' was missing from the integration since I used polar coordinate.